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This approach employs DFS for exploring each possible path starting from the given start index. You recursively explore both forward and backward jumps, marking indices as visited to avoid cycles. If you reach an index with a value of 0, return true. If all paths are exhausted without finding a zero, return false.
Time Complexity: O(n), where n is the length of the array, because every index is visited at most once.
Space Complexity: O(n) due to the recursion stack and the visited array.
1var canReach = function(arr, start) {
2 const dfs = (arr, pos) => {
3 if (pos < 0 || pos >= arr.length || arr[pos] < 0)
4 return false;
5 if (arr[pos] === 0)
6 return true;
7 const jump = arr[pos];
8 arr[pos] = -arr[pos]; // Mark visited
9 const result = dfs(arr, pos + jump) || dfs(arr, pos - jump);
10 arr[pos] = -arr[pos]; // Reset
11 return result;
12 };
13 return dfs(arr, start);
14};The JavaScript solution implements a similar approach using a DFS methodology. Recursion checks for boundary conditions and zero values while marking visited indices with negative values as a simple visited marker.
This approach adopts a BFS strategy to explore all reachable indices from the starting point iteratively. Using a queue, enqueue all possible jump indices from a given position if they haven't been visited yet. Pop indices from the queue, and for each pop, check whether the position contains a zero, marking indices visited as you go.
Time Complexity: O(n), as all reachable indices are visited.
Space Complexity: O(n), given the storage requirements for the queue and the visited array.
1
Here's how the C solution works: A queue is dynamically created and used to explore all potential positions, marking indices as visited in a dedicated boolean array. By processing each queue element, we're checking both forward and backward jumps. This approach terminates when an index with a zero value is found or when the queue is empty.